## Essential University Physics: Volume 1 (3rd Edition)

We first find the maximum allowable velocity: $6 = \frac{mv^2}{r} + mg$ $6 = \frac{(.5)v^2}{.45} + (.5)(9.81)$ $v=.9927 \ m/s$ Thus, we find the height that it can go: $mgh = \frac{1}{2}mv^2 \\ h=\frac{v^2}{2g}=\frac{.9927^2}{2\times9.81}=.05$ Thus, we use trigonometry to find: $cos\theta = \frac{.4}{.45} \\ \theta = 27.26^{\circ}$